Problem: You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

* 2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:


N


Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Example

Input

2


Output

4 7
23 10

Solution (in CPP):

Solution:
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N=5e2+5;
const int M=5e5+5;
int n,cnt,prime[M],vis[M];
ll a[N][N];

ll gcd(ll x,ll y) {return y?gcd(y,x%y):x;}
ll lcm(ll x,ll y) {if(!x||!y) return x+y;return 1ll*x/gcd(x,y)*y;}
template <typename _Tp> inline void IN(_Tp&x) {
	char ch;bool flag=0;x=0;
	while(ch=getchar(),!isdigit(ch)) if(ch=='-') flag=1;
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	if(flag) x=-x;
}

int main() {
	IN(n);
    if(n==2) return printf("4 7\n23 10\n"),0;
	for(int i=2;i<M;++i) {
		if(!vis[i]) prime[++cnt]=i;
		for(int j=1;i*prime[j]<M&&j<=cnt;++j) {
			vis[i*prime[j]]=1;
			if(!(i%prime[j])) break;
		}
	}
    for(int i=1;i<=n;++i)
        for(int j=(i+1&1)+1;j<=n;j+=2)
            a[i][j]=prime[(i+j)/2]*prime[n+(i-j)/2+(n+1)/2];
    for(int i=1;i<=n;++i)
        for(int j=(i&1)+1;j<=n;j+=2)
            a[i][j]=lcm(lcm(a[i-1][j],a[i][j-1]),lcm(a[i+1][j],a[i][j+1]))+1;
	for(int i=1;i<=n;++i) {
        for(int j=1;j<=n;++j) printf("%lld ",a[i][j]);
        puts("");
    }
	return 0;
}